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How do I interpret the result:

ARIMAProcess[2.22836, {0.363326}, 1, {}, 6.35083]

in terms of the equivalent recurrence equation; say

y[t] = 2.22836 + 0.363326y[t-1]...?

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Perhaps you can use ARProcess representation of ARIMAProcess[2.22836, {0.363326}, 1, {}, 6.35083]:

proc = ARIMAProcess[c, {a}, 1, {}, v];
arproc = ARProcess[proc, 5]

ARProcess[c, {1 + a, -a}, v]

Using the general difference equation representation from ARProcess >> Details

Mathematica graphics

we get, for aproc:

Mathematica graphics

where e(t) is white noise with variance v.

Alternatively,

 y[t] = c + (1+a) y[t-1] - a y[t-2] + Sqrt[v] u[t]

where u[t] is white noise with variance 1.

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  • $\begingroup$ Many thanks for your reply ... prg $\endgroup$ – PRG Dec 20 '16 at 15:54
  • $\begingroup$ @PRG, my pleasure. Thank you for the accept. Welcome to mma.se. $\endgroup$ – kglr Dec 20 '16 at 21:04

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